Problem: Multiply the following rational expressions and simplify the result. $\dfrac{9y-33y^2-3y^4}{100-49y^2} \cdot \dfrac{7y^2+17y+10}{14y^2+28y}=$
Answer: Let's first factor the numerators and denominators of each expression separately. [Why are we doing this?] The numerator, $9y-33y^2-3y^4$, of the first expression can be factored as $-3y(y^3+11y-3)$ by factoring out $-3y$. The denominator, $100-49y^2$, of the first expression can be factored as $(10+7y)(10-7y)$ using the difference of squares pattern. The numerator, $7y^2+17y+10$, of the second expression can be factored by grouping as $(y+1)(7y+10)$. The denominator, $14y^2+28y$, of the second expression can be factored as $14y(y+2)$ by factoring out $14y$. Now the product looks as follows: $\dfrac{-3y(y^3+11y-3)}{(10+7y)(10-7y)}\cdot\dfrac{(y+1)(7y+10)}{14y(y+2)}$ To find the product of two rational expressions, we multiply across, then simplify: [What's that?] $\phantom{=}\dfrac{-3y(y^3+11y-3)}{(10+7y)(10-7y)}\cdot\dfrac{(y+1)(7y+10)}{14y(y+2)}$ $\begin{aligned} &= \dfrac{-3y(y^3\!+\!11y\!-\!3) \cdot (y\!+\!1)(7y\!+\!10)}{(10\!+\!7y)(10\!-\!7y) \cdot 14y(y\!+\!2)} &\text{Multiply across.}\\\\\\ &= \dfrac{-3{\cancel{y}}(y^3\!+\!11y\!-\!3) (y\!+\!1){\cancel{(7y\!+\!10)}}}{{\cancel{(10\!+\!7y)}}(10\!-\!7y) 14{\cancel{y}}(y\!+\!2)} &\text{Cancel out common factors.}\\\\\\\\ &=\dfrac{-3(y^3+11y-3)(y+1)}{14(10-7y)(y+2)} \end{aligned}$ Therefore, the simplified form of the product is $\dfrac{-3(y^3+11y-3)(y+1)}{14(10-7y)(y+2)}$, which is equivalent to $\dfrac{3y^4+3y^3+33y^2+24y-9}{98y^2+56y-280}$.